Đáp án:
Giải thích các bước giải:
Đặt $A=$ $\frac{1}{10}$+ $\frac{1}{15}$+ $\frac{1}{21}$+...+ $\frac{1}{120}$
⇒$\frac{1}{2}$$A=$$\frac{1}{20}$+ $\frac{1}{30}$+ $\frac{1}{42}$+...+ $\frac{1}{240}$
⇔$\frac{1}{2}$$A=$$\frac{1}{4.5}$+ $\frac{1}{5.6}$+ $\frac{1}{6.7}$+...+ $\frac{1}{15.16}$
⇔$\frac{1}{2}$$A=$$\frac{1}{4}$- $\frac{1}{5}$+ $\frac{1}{5}$- $\frac{1}{6}$+ $\frac{1}{6}$- $\frac{1}{7}$+...+ $\frac{1}{15}$- $\frac{1}{16}$
⇔$\frac{1}{2}$$A=$ $\frac{1}{4}$-$\frac{1}{16}$=$\frac{12}{64}$= $\frac{3}{16}$
⇒$A= $$\frac{3}{8}$
